AP EAMCET · PHYSICS · Motion In Two Dimensions
A particle is projected at an angle of \(60^{\circ}\) with the horizontal from the ground with a velocity \(10 \sqrt{3} \mathrm{~ms}^{-1}\). The angle between velocity vector after \(2 \mathrm{~s}\) and initial velocity vector is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A 0
- B \(30^{\circ}\)
- C \(60^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(D) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
Initial velocity, \(\mathbf{v}_{\mathbf{i}}=2 \cos \theta \hat{\mathbf{i}}+4 \sin \theta \hat{\mathbf{j}}=5 \sqrt{3} \hat{\mathbf{i}}+15 \hat{\mathbf{j}}\) Final velocity vector (after \(2 s\) ),…
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